# Calculus PLEASE check my work ,

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1.) which of the following represents dy/dx when y=e^-2x Sec(3x)?
A.3e^-2x sec(3x) tan (3x)-2e^-2x Sec(3x)<<<< my choice
B.)3e^-2x sec(3x)tan (3x)-2xe^-2x sec(3x)
C.)3e^-2x sec(3x)tan (x)-2e^-2x Sec(3x)
D.)3e^-2x sec(3x)tan (x)-2xe^-2x sec(3x)

2.)calculate dy/dx if y=Ln(2x^3+3x)
A.)1/2x^3+3x
B.)1/6x^2+3
C.)2x^3+3x/6x^2+3
D.)6x^2+3/2x^3+3x <<<< my choice

3.)Given the curve that is described by the equation r=3 cos theta, find the angle that tangent lines makes with the radius vector when theta=120.
A.) 30deg
B.) 45deg
C.) 60deg
D.) 90deg << my choice

4.)A line rotates in a horizontal plane according to the equation theta=2^t-6t,where theta is the angular position of the rotating line, in radians ,and tis the time,in seconds. Determine the angular acceleration when t=2sec.

D.) 24 radians per sec^2 << my choice.

• Calculus PLEASE check my work , - ,

1. Correct
2. Correct
3.
dy/dx
=(dy/dt) / (dx/dt)
put
x=rcos(t)=3cos²(t)=
y=rsin(t)=3sin(t)cos(t)
Calculate dy/dx and evaluate at t=120° (2π/3) to get
dy/dx=-1/√3
=> θ=150°

4.
There may have been a typo in the question. If the question is asking for θ=2^t-6t, the answers should not be in round numbers.

• Calculus PLEASE check my work , - ,

A
line
rotates
in
a horizontal
plane
according
to
the
equation
9
=
2t3-
6t,
where
(J
is
the
angular
position
of
the
rotating
line,
in
and
t
is the
time,
in
sec-
onds.
Determine
the
angular
acceleration
when
t
=
2
sec.